Question

A cylindrical mild steel tensile test specimen of gauge length 50 mm and diameter 10 mm is extended in two stages at a deformation speed of 4 mm/min. The specimen is extended from 50 mm to 55 mm in the first stage, and from 55 mm to 60 mm in the second stage. Neglecting elastic deformation, the total longitudinal true strain is _________.
\text{[round off to 2 decimal places]}

Hint:

For total true strain, sum the true strains from each stage of deformation.

Answer 1

Correct Answer: 0.17

The true strain is given by: \[ \epsilon_{\text{true}} = \ln \left( \frac{L_{\text{final}}}{L_{\text{initial}}} \right). \] For the first stage, the true strain is: \[ \epsilon_1 = \ln \left( \frac{55}{50} \right) = \ln(1.1) \approx 0.0953. \] For the second stage: \[ \epsilon_2 = \ln \left( \frac{60}{55} \right) = \ln(1.0909) \approx 0.0870. \] The total true strain is: \[ \epsilon_{\text{total}} = \epsilon_1 + \epsilon_2 = 0.0953 + 0.0870 = 0.1823. \] Thus, the total longitudinal true strain is approximately \( 0.17 \).